Protocol Tome Tracker 4 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a filtering and aggregation routine for a sequence of integer metrics. Given an array of integers representing data points and a threshold integer K, your objective is to compute the cumulative sum of all elements that strictly exceed the threshold K. Elements equal to or less than K must be ignored in the final calculation. If no elements satisfy the condition, the result is zero.
The input consists of a single array of integers and an integer K. The output is a single integer representing the sum of the qualifying elements. This problem tests your ability to iterate through a collection and apply a conditional filter before performing an arithmetic reduction.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Tome Tracker 4"
WHY DOES IT MATTER?
Filtering‑then‑aggregating is a fundamental pattern because many real‑world metrics require summarizing data that meets a simple predicate. Mastering this pattern prevents over‑engineering solutions that waste time and resources.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the predicate is independent per element, allowing a single linear scan instead of exploring combinatorial subsets. This eliminates exponential backtracking and reduces both time and space to their minimal bounds.
REAL-WORLD CONNECTION
Think of a log‑processing pipeline that only sums request latencies exceeding a SLA threshold. The pipeline streams logs, discards sub‑threshold entries, and aggregates the rest—mirroring the algorithm's filter‑and‑sum behavior.
During an interview, write the loop first, then immediately add the condition if (arr[i] > K) sum += arr[i];. This shows you understand the greedy approach and avoids unnecessary data structures.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem asks for the sum of all array elements that are strictly greater than a given threshold K. The most direct way to solve this is to iterate through the array once, checking each element against K and accumulating the qualifying values. This linear scan leverages the fact that the operation required for each element (a comparison and a possible addition) is O(1), leading to an overall O(N) time solution where N is the length of the array.
A naive approach might attempt to generate all possible subsets of the array and then sum each subset to see if it contains only elements > K. Such a power‑set enumeration has exponential complexity (O(2^N)) and quickly becomes infeasible even for modest N. The exponential blow‑up occurs because backtracking explores every combination, which is unnecessary when the condition applies independently to each element. Recognizing that the predicate "element > K" is separable allows us to discard the combinatorial explosion and adopt a simple greedy scan.
The optimal paradigm here is a single‑pass greedy aggregation. Since the decision for each element does not depend on any other element, we can process the array in any order, maintain a running total, and finish in linear time with constant extra space. This pattern—filter‑then‑aggregate—is a staple in algorithm design for problems that involve simple thresholds or predicates.
Interview Questions on This Problem
Q1How would you modify the solution if you needed to return the count of elements greater than K in addition to their sum?
Maintain two variables during the single pass: one for the sum and another for the count. Increment the count each time an element > K is encountered and add the element to the sum. Both operations remain O(1) per element, preserving O(N) overall time and O(1) space.
Q2Can you solve the problem in a functional programming style using built‑in language constructs?
Yes. In languages like Python, JavaScript, or Java Streams, you can filter the array with a predicate (x > K) and then reduce the filtered stream by summing. For example, Python: sum(x for x in arr if x > K). This abstracts the loop but still runs in O(N) time and O(1) auxiliary space.
Q3If the input array is extremely large and stored on disk, how would you adapt the algorithm to handle it efficiently?
Process the data in a streaming fashion: read chunks of the array sequentially, apply the same >K check, and update a running total. This avoids loading the entire array into memory, keeping memory usage O(1) while still achieving O(N) time over the total number of elements.
Examples
Input
nums = [12, 5, 8, 15, 3], K = 10
Output
27
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 8 <= 10 (skip), 15 > 10 (add 15), 3 <= 10 (skip). The sum is 12 + 15 = 27.
Input
nums = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: Iterate through the array: All elements (1, 2, 3, 4, 5) are less than or equal to 10. No elements are added to the sum. The final result is 0.
Input
nums = [10, 10, 10, 11], K = 10
Output
11
Explanation: Iterate through the array: The first three elements are equal to 10, so they are skipped because the condition is strictly greater than. The last element 11 is greater than 10, so it is added. The sum is 11.
Input
nums = [-5, -2, 0, 3, 7], K = 0
Output
10
Explanation: Iterate through the array: -5 <= 0 (skip), -2 <= 0 (skip), 0 <= 0 (skip), 3 > 0 (add 3), 7 > 0 (add 7). The sum is 3 + 7 = 10.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Perform a single linear scan, adding each element to the answer only if it exceeds K. This yields O(N) time and O(1) auxiliary space.
Brute Force Approach
Generate every subset of the array, compute the sum of each subset, and keep the maximum sum of subsets where all elements are > K. This is exponential and impractical.
Verified Code Solutions
function solution(nums, K) { return nums.filter(num => num > K).reduce((a, b) => a + b, 0); }class Solution { public: int solution(vector<int>& nums, int K) { int sum = 0; for (int num : nums) { if (num > K) { sum += num; } } return sum; } };class Solution { public int solution(int[] nums, int K) { int sum = 0; for (int num : nums) { if (num > K) { sum += num; } } return sum; } }def solution(nums, K): return sum(num for num in nums if num > K)function solution(nums, K) { return nums.filter(num => num > K).reduce((a, b) => a + b, 0); }Asked in Top Tech Interviews
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